Wave function of a particle in a potential barrier In our course we considered several examples of a particle in a potential wall. One case was that of a particle moving along the $x$ axis and the potential was a step function of the form:
$x<0 $: $V(x)=0$
$x>0 $: $ V(x)=V_0$
So we divide the region in two parts: Region 1: $x<0$ and Region 2: $x>0$
The wave function in the first region (x<0) was given as:
$\psi_1=A_1e^{ik_1x}+B_1e^{-ik_1x}$
$A_1e^{ik_1x}$ represents the incident wave.
$B_1e^{-ik_1x}$ represents the reflecting wave (on the potential wall).
In the 2 region (x>0):
$\psi_2=A_2e^{ik_2x}+B_2e^{-ik_2x}$
We consider that $B_2e^{-ik_2x}=0 \rightarrow B_2=0$, in other words there is not wave coming from the right side.
$A_2e^{ik_2x}$ is the transmitted wave.
While I understand, logically what is happening here for a matter wave, is similar to a electromagnetic wave/light wave when it passes throught mediums with different refracting index, two things I don't understand are:

*

*Where was this wave function expression $\psi_1=A_1e^{ik_1x}+B_1e^{-ik_1x}$ derived from?


*Is the state $\psi_1$ (or $\psi_2$ for that matter) of the system (particle) a superposition of  eigenstates of a certain operator, or an eigenstate of an operator. Whether that Operator can or cannot be the Hamilton operator
 A: Merely answering:

I have searched for a long time, for a derivation of the solution to
the wave eq. in the form $u(x,t)=u_0e^{i(kx−\omega t)}$ but I haven't found one
yet. Can you help me with one?

by OP in the in the comment section. You may have 'searched for a long time' because there's no solution of that form.
For a particle in $\text{1D}$ the Time Dependent Schrödinger Equation is (and where I'm using $\Psi$ instead of the OP's $u$):
$$i\hbar \partial_t \Psi(x,t)=\hat{H}\Psi(x,t)\tag{1}$$
where :
$$\hat{H}=-\frac{\hbar^2}{2m}\partial_{xx}+V(x)$$
Assume ('Ansatz'):
$$\Psi(x,t)=\psi(x)\phi(t)\tag{2}$$
Separation of variables: insert $(2)$ into $(1)$
$$i\hbar \psi \phi'=-\frac{\hbar^2}{2m}\psi'' \phi+\psi \phi V(x)$$
Divide by $\psi \phi$:
$$i\hbar\frac{\phi'}{\phi}=-\frac{\hbar^2}{2m}\frac{\psi''}{\psi}+V(x)=E$$
where $E$ is a separation constant (total energy in this case). We get $2$ ODEs:
$$i\hbar\frac{\phi'}{\phi}=E\tag{3}$$
and after minimal reworking:
$$\hat{H}\psi(x)=E\psi(x)\tag{4}$$
$(4)$ is of course the Time Independent Schrödinger Equation, while $(3)$ solves to:
$$\phi(t)=e^{-iEt/\hbar}$$
So $\Psi(x,t)$ takes the form:
$$\Psi(x,t)=\psi(x)e^{-iEt/\hbar}$$
The step potential of OP's question, $\psi(x)$ is derived here.
A: To answer your first question, the equation you have to solve is the Schrödinger equation (nothing suprising) with the potential $V(x) = \delta(x)$. This makes sense as this means that there is no potential for $x < 0$ nor for $x > 0$ and leaves us with an infinite barrier at $ x = 0 $. This is where you get the solving of two different wave functions. Both equations have the following form (no time dependence here):
$
-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} = E \psi
$
while keeping in mind that one is valid for negative positions and the other one for positive ones. It should be clear that the solution to this is a superposition of waves, one with positive and one with negative frequency. Then the next step is to impose continuity and derivability of the full solution, which implies that $\psi_1$ and $\psi_2$ must have the same values and the same derivative at $x = 0$. The last step would be to impose boundary conditions such as $B = 0$.
Now we are in position to answer your second question. $\psi_1$ and $\psi_2$ are themselves eigenstates, as should be clear from above. I understand that you are familiar with the free particle solution to the Schrödinger equation, and know that the energy eigenvalues are continious. The same applies here: there are two regions of free particle solutions so the full solution will be a continious superposition of all states (usually this is very messy and it is done by computer). This of course means that the operator giving rise to the eigenstates is the Hamiltonian operator, which also should be clear from above.
Please keep in mind this is only for the case you presented above, which implies that E > 0. For bound states (i.e. E < 0), there is only one posible state in the delta function potential.
